MATH201/content/Power series (lec 22-23).md

#start of lec 22
Finished chapter 7 of the course textbook, Let's begin chapter 8!
# Power series
A power series is defined by:
$$\sum_{n=0}^\infty a_{n}(X-X_{0})^n=a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})^2+\dots$$
It is convergent if:
$$\sum_{n=0} ^ \infty a_{n}(x-x_{0})^n<\infty \text{ at a given x}$$
Otherwise, it is divergent.
If $\sum_{n=0}^\infty \mid a_{n}(x-x_{0})^n\mid$ is convergent
 $\implies\sum_{n=0}^\infty  a_{n}(x-x_{0})^n$ is absolutely convergent
Just because something is absolutely convergent doesn't mean it is conditionally convergent. think of the harmonic series. It is absolutely convergent but also divergent (conditionally divergent).

Theorem: With each $\sum_{n=0}^{\infty}a_{n}(x-x_{0})^n$ we can associate $0\leq \rho\leq \infty$ such that
$\sum_{n=0} ^\infty a_{n}(x-x_{0})^n$ is absolutely convergent 
for all x such that $\mid x-x_{0}\mid<\rho$, divergent for all x where $\mid x-x_{0}\mid>\rho$
"Who keeps stealing the whiteboard erases? (jokingly) It's a useless object, anyways"
![[Drawing 2023-10-30 13.12.57.excalidraw.png]]
how can we find $\rho$?
Ratio test: If $\lim_{ n \to \infty }\mid \frac{a_{n+1}}{a_{n}}\mid=L$
then $\rho=\frac{1}{L}$

## Examples:
#ex 
is this convergent? Divergent? and where so?
$\sum_{n=0}^\infty \frac{2^{-n}}{n+1}(x-1)^n$
determine the convergent set.
Use ratio test:
$\lim_{ n \to \infty }\mid \frac{a_{n+1}}{a_{n}}\mid=\lim_{ n \to \infty } \frac{2^{-(n+1)}}{n+2} \frac{n+1}{2^{-n}}=\frac{1}{2}\implies \rho=2$
so it's convergent on $-1<x<3$, divergent on $\mid x-1\mid>2$
But what about on the points $-1$ and $3$?
plug in $x_{0}=-1$
$\sum_{n=0}^{\infty} \frac{2^{-n}}{n+1}(-2)^n=\sum_{n=0}^\infty \frac{(-1)^n}{n+1}<\infty$ <- That is the alternating harmonic series, it is convergent.
plug in $x_{0}=3$:
$\sum_{n=0}^\infty \frac{2^{-n}}{n+1}2^n=\sum_{n=0}^\infty \frac{1}{n+1}>\infty$ <- harmonic series, this diverges.
so the power series is convergent on $[-1,3)$ divergent otherwise.
$$\text{ converges only on: } [-1,3)$$

Assume that $\sum_{n=0}^\infty  a_{n}(x-x_{0})^n$ and $\sum_{n=0}^\infty  b_{n}(x-x_{0})^n$ are converget with $\rho>0$
Then: 
1.) $\sum_{n=0}^\infty a_{n}(x-x_{0})^n+\sum_{n=0}^{\infty}b_{n}(x-x_{0})^n=\sum_{n=0}^\infty(a_{n}+b_{n})(x-x_{0})^n$
That has a radius of convergence of at least $\rho$.
2.) $\left( \sum_{n=0}^\infty a_{n}(x-x_{0})^n \right)\left( \sum_{n=0}^\infty b_{n}(x-x_{0})^n \right) \qquad c_n=\sum_{k=0}^n a_{k}b_{n-k}$(Cauchy)
$=(a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})^2+\dots)(b_{0}+b_{1}(x-x_{0})+b_{2}(x-x_{0})^2+\dots)$
$=a_{0}b_{0}+(a_{0}b_{1}+a_{1}b_{0})(x-x_{0})+(a_{0}b_{2}+a_{1}b_{1}+a_{2}b_{1})(x-x_{0})^2+\dots$ (Cauchy multiplication)

more Definitions of power series:
If $\sum_{n=0}^{\infty}a_{n}(x-x_{0})^n$ is convergent with $\rho>0$
$\mid x-x_{0}\mid<\rho$
we can differentiate this infinite sum and get:
$\implies y'(x)=\sum_{n=1}^\infty a_{n}n(x-x_{0})^{n-1}$
$y''(x)=\sum_{n=2}^\infty a_{n}n(n-1)(x-x_{0})^{n-2}$

Theorem: If $y(x)$ is infinitely many times differentiable on some interval: $\mid x-x_{0}\mid<\rho$
then: $\sum_{n=0}^\infty \frac{y^{(n)}(x_{0})}{n!}(x-x_{0})^n$ (Taylor series)
"believe me, taylor series is the most important theorem in engineering."
"I mean engineering is all about approximations, do you know how your calculator computes ...? Taylor series!"
"Applied mathematics is all about approximating and then measuring how good your approximation is, it's what engineering is all about." -Prof (loosy quotes, can't keep up with how enthusiastic he is!)

Definition: If $y(x)$ can be represented with a power series on $\mid x-x_{0}\mid$ then $y(x)$ is an analytic function on $(x_{0}-\rho,x_{0}+\rho)$
btw analytic functions are very important in complex calculus MATH301. (i don't have that next term)

$f(x)=\sum_{n=0}^\infty a_{n}(x-x_{0})^n<\infty$
$f'(x)=\sum_{n=1}^\infty a_{n}n(x-x_{0})^{n-1}$
$f(x)+f'(x)=\sum_{n=0}^\infty a_{n}(x-x_{0})^n+\sum_{n=1}^\infty a_{n}n(x-x_{0})^{n-1}$
let $n-1=k$
$=\sum_{n=0}^\infty a_{n}(x-x_{0})^n+\sum_{k=0}^\infty a_{n}(k+1)(x-x_{0})^{k}$
$=\sum_{n=0}^\infty(a_{n}+a_{n}(n+1))(x-x_{0})^n$

Last theorem fo' da day:
If $\sum_{n=0}^\infty a_{n}(x-x_{0})^n=0$ for all x$\in(x_{0}-\rho,x_{0}+\rho)$ where $\rho>0$
$\implies a_{n}=0$, $n=0,1,2,\dots$
#end of lec 22 #start of lec 23
Mid terms are almost done being marked!

## Examples
Let's start using power series to start solving DE!
No magic formulas we need to memorize when solving equations using power series (Yay!)
#ex 
$$y'-2xy=0 \qquad x_{0}=0$$
note this is separable and linear, so we can already solve this. This time we do it with power series
y should be an analytic function (meaning, infinitely many times differentiable)
so we should expect we can represent it as a power series
$y(x)=\sum_{n=0}^\infty a_{n}x^n$
$y'(x)=\sum_{n=1}^\infty a_{n}nx^{n-1}$
plug these into the equation:
$\sum_{n=1}^\infty a_{n}nx^{n-1}-\sum_{n=0}^\infty 2a_{n}x^{n+1}=0$
if the entire interval is zero, we should expect all the coefficients to equal 0
we need to combine the summations.
shift the index!
$k=n-1,\ k=n+1$
$\sum_{k=0}^\infty a_{k+1}(k+1)x^{k}-\sum_{k=1}^\infty 2a_{k-1}x^k=0$


$a_{1}+\sum_{k=1}^\infty (\underbrace{ a_{k+1}(k+1)-2a_{k-1} }_{ =0 })x^k=0$
The whole series equals zerro,
so $a_{1}=0$ is the first observation
second observation:
$a_{k+1}=\frac{2}{k+1}a_{k-1}$ where $k=1,2,3,\dots$ This is called a recursive relation. (if we know one index we can produce some other index recursively)
from this equation:
$a_{1}, a_{3}, a_{5}, \dots=0$
$a_{2k+1}=0, k=0,1,2,\dots$
this means half of our power series disappears!
what happens with the other half?
$a_{2}$ is related to $a_{0}$ from the above formula
$a_{2}=\frac{2}{2}a_{0}$ ($k=1$)
$a_{4}=\frac{2}{3+1}a_{2}=\frac{a_{0}}{2}$ ($k=2$)
$a_{6}=\frac{2}{5+1} \frac{a_{0}}{2}=\frac{a_{0}}{6}$ ($k=3$)
$a_{8}=\frac{1}{4} \frac{a_{0}}{6}=\frac{a_{0}}{24}$ ($k=4$)
you might start noticing a factorial-y pattern:
$a_{2k}=\frac{1}{k!}a_{0}$ where $k=0,1,2,\dots$

$y(x)=a_{0}\sum_{k=0} ^\infty \frac{1}{k!}x^{2k}=a_{0}\sum_{k=0} ^\infty \frac{1}{k!}(x^2)^k$
Does this look like something from math 101?
Yes! it looks like the taylor series of $e^{x^2}$
so: 
$$y(x)=a_{0}e^{x^2}$$
"if we are correct--the same is not true in general in real life--but in mathematics if we are correct we should end up with the same solution" -Prof
#ex 
$$z''-x^2z'-xz=0 \qquad \text{about } x_{0}=0$$
using regular methods will be problematic,
if you use laplace transform you will have problems as well.
>"you try the simplest thing you know, if you know anything :D" (referring to answering a question about how do we know what method to use?)

lets use power series:
assume solution is analytic:
$z(x)=\sum_{n=0}^\infty a_{n}nx^{n-1}$
$z''(x)=\sum_{n=2}^\infty a_{n}n(n-1)x^{n-2}$
$\underset{ n-2=k }{ \sum_{n=2}^\infty a_{n}n(n-1)x^{n-2} }-\underset{ n+1=k }{ \sum_{n=1}^\infty a_{n} nx^{n+1} }-\underset{ n+1=k }{ \sum_{n=0}^\infty a_{n}x^{n+1} }=0$

shift the index to equalize the powers.
>loud clash of clans log in sound, class giggles, "whats so funny?" :D "im not a dictator" something about you are not forced to sit through and watch the lecture if you don't like to, "I dont think everybody should like me."

$\sum_{k=0}^\infty a_{k+2}(k+2)(k+1)x^k-\sum_{k=2}^\infty a_{k-1}x^k-\sum_{k=1}^\infty a_{k-1}x^k=0$
$2a_{2}+6a_{3}+\sum_{k=2}^\infty a_{k+2}(k+2)(k+1)x^k-\sum_{k=2}^\infty a_{k-1}(k-1)x^k-a_{0}x-\sum_{k=2}^\infty a_{k-1}x^k=0$
$6a_{2}+(6a_{3}-a_{0})x+\sum_{k=2}^\infty (a_{k+2}(k+1)(k+2)-a_{k-1}k)x^k=0$
$a_{2}=0 \qquad a_{3}=\frac{a_{0}}{6}$
$a_{k+2}=\frac{k}{(k+1)(k+2)}a_{k-1}$ where $k=2,3,4,\dots$
Finally, a recursive relation!
it should be clear that each step of 3 starting from $a_{2}$ should all equal 0.
$a_{2}, a_{5}, a_{8}, \dots=0$
$a_{3k-1}=0$ where $k=1,2,\dots$
$a_{4}=\frac{2}{3\cdot 4}a_{1}$
$a_{7}=\frac{5}{6\cdot 7} \frac{2}{3\cdot 4}a_{1}$
realize if we multiply here by 5 and 2:
$a_{7}=\frac{5^2}{5\cdot6\cdot 7} \frac{2^2}{2\cdot3\cdot 4}a_{1}$
$a_{7}=\frac{(2\\dot{c} 5)^2}{7!}a_{1}$
$a_{4}=\frac{2^2}{4!}a_{1}$
the pattern leads us to:
$a_{3k+1}=\frac{(2\cdot 5 \dots(3k-1))^2}{(3k+1)!}$ where k=1,2,3, ...
$a_{3k}=\frac{(1\cdot 4\cdot \dots(3k-2))^2}{(3k)!}a_{0}$ k=1,2,...
$z(x)=a_{0}\left( 1+\sum_{k=1}^\infty \frac{(1*4*\dots(3k-2))^2}{(3k)!}x^{3k} \right)$
$a_{1}\left( x+\sum_{k=1}^\infty \frac{(2\cdot 5\cdot \dots(3k-1))^2}{(3k+1)!} x^{3k+1}\right)$
there we go, $z$ is a linear combination of those two expressions
class done at 1:56 (a lil late but the journey is worth it)
#end of lec 23